电力系统潮流短路 计算课程设计(5)
发布时间:2021-06-06
发布时间:2021-06-06
S1 1 =
′SLD14
+
SLD14′2UNB22
R2+X2 jUN
= 4.217+j1.197 + 0.0333 j1.6305 MVA= 4.2503 j0.433 MVA
⑥ 1.3间120Km线路损耗 SLD13
1010
5.651+j5.154 = 2.569+j2.343 MVA =S3=
B12
jUN= 2.569+j0.3102 MVA
SLD13′2UNB12
R1+X1 jUN
SLD13′=SLD13S1 2 =
′SLD13
+
= 2.596+j0.3102 + 0.0139+j0.0266 j2.0328MVA= 2.61 j1.696 MVA
⑦ 2.4间70Km线路损耗 SLD24
1010
10.2405+j7.02 = 6.0238+j4.1294 MVA =S4=
B42
jUN= 6.0238+j2.9436 MVA
SLD24′2UNB42
R4+X4 jUN
SLD24′=SLD24
S2 1 =SLD24′+
= 6.0238+j2.9436 + 0.0546+j0.104 j1.1856MVA= 6.0784+j1.862 MVA
⑧ 2.3间100Km线路损耗 SLD24
1212
5.651+j5.154 = 3.082+j2.811 MVA =S=
4B32
jUN= 3.082+j1.1172 MVA
SLD23′=SLD23
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